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PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway.

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Presentation on theme: "PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway."— Presentation transcript:

1 PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway

2 Tuning of PID controllers SIMC tuning rules (“Skogestad IMC”) (*) Main message: Can usually do much better by taking a systematic approach Key: Look at initial part of step response Initial slope: k’ = k/  1 One tuning rule! Easily memorized Reference: S. Skogestad, “Simple analytic rules for model reduction and PID controller design”, J.Proc.Control, Vol. 13, 291-309, 2003 (*) “Probably the best simple PID tuning rules in the world”  c ¸ 0: desired closed-loop response time (tuning parameter) For robustness select:  c ¸ 

3 Need a model for tuning Model: Dynamic effect of change in input u (MV) on output y (CV) First-order + delay model for PI-control Second-order model for PID-control

4 Step response experiment Make step change in one u (MV) at a time Record the output (s) y (CV)

5 First-order plus delay process Step response experiment k’=k/  1 STEP IN INPUT u (MV) RESULTING OUTPUT y (CV)  Delay - Time where output does not change  1 : Time constant - Additional time to reach 63% of final change k : steady-state gain =  y(1)/  u k’ : slope after response “takes off” = k/  1

6 Model simplifications For control: Dynamics at “control time scale” (  c ) are important Integrating process (  1 = 1 ) Time constant  1 is not important if it is much larger than the desired response time  c. More precisely, may use  1 = 1 for  1 > 5  c Delay-free process (  =0) Delay  is not important if it is much smaller than the desired response time  c. More precisely, may use  ¼ 0 for  <  c /5  ¼ 1 (may be neglected for  c > 5)  1 ¼ 200 (may be neglected for  c < 40) time  c = desired response time

7 “Integrating process” (  c < 5  1 ):  Need only two parameters: k’ and   From step response: Response on stage 70 to step in L 7.5 min 2.62-2.19  Example. Step change in u:  u = 0.1 Initial value for y: y(0) = 2.19 Observed delay:  = 2.5 min At T=10 min: y(T)=2.62 Initial slope: y(t) t [min]

8 First-order with delay process (  c > 5  1 ) Need 3 parameters:  Delay   2 of the following: k, k’,  1 (Note relationship k’ = k/  1) Example: First-order model:  = 0 k= (0.134-0.1)/(-0.5) = -0.068 (steady-state gain)  1 = 16 min (63% of change)  Gives k’ = 0.068/16 = 0.00425  (could alternatively find k’ by observing the initial response)  BOTTOM V: -0.5 xBxB 16 min 63%

9 Step response experiment: How long do we need to wait? FAST TUNING DESIRED (“tight control”,  c =  ):  NORMALLY NO NEED TO RUN THE STEP EXPERIMENT FOR LONGER THAN ABOUT 10 TIMES THE EFFECTIVE DELAY (  )  EXCEPTION: LET IT RUN A LITTLE LONGER IF YOU SEE THAT IT IS ALMOST SETTLING (TO GET  1 RIGHT)  SIMC RULE:   = min (  , 4(  c +  )) with  c =  for tight control SLOW TUNING DESIRED (“smooth control”,  c >  ):  HERE YOU MAY WANT TO WAIT LONGER TO GET  1 RIGHT BECAUSE IT MAY AFFECT THE INTEGRAL TIME  BUT THEN ON THE OTHER HAND, GETTING THE RIGHT INTEGRAL TIME IS NOT ESSENTIAL FOR SLOW TUNING  SO ALSO HERE YOU MAY STOP AT 10 TIMES THE EFFECTIVE DELAY (  )

10 Model reduction of more complicated model Start with complicated stable model on the form Want to get a simplified model on the form Most important parameter is usually the “effective” delay 

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13 half rule

14 Approximation of zeros

15 Derivation of SIMC-PID tuning rules PI-controller (based on first-order model) For second-order model add D-action. For our purposes it becomes simplest with the “series” (cascade) PID-form:

16 Basis: Direct synthesis (IMC) Closed-loop response to setpoint change Idea: Specify desired response (y/y s )=T and from this get the controller. Algebra:

17 IMC Tuning = Direct Synthesis

18 Integral time Found: Integral time = dominant time constant (  I =  1 ) Works well for setpoint changes Needs to be modified (reduced) for integrating disturbances Example. “Almost-integrating process” with disturbance at input: G(s) = e -s /(30s+1) Original integral time  I = 30 gives poor disturbance response Try reducing it! gc d y u

19 Integral Time  I =  1 Reduce  I to this value:  I = 4 (  c +  ) = 8 

20 SIMC-PID Tuning Rules One tuning parameter:  c

21 Some special cases One tuning parameter:  c

22 Note: Derivative action is commonly used for temperature control loops. Select  D equal to time constant of temperature sensor

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24 Selection of tuning parameter  c Two cases 1. Tight control: Want “fastest possible control” subject to having good robustness 2. Smooth control: Want “slowest possible control” subject to having acceptable disturbance rejection

25 TIGHT CONTROL

26 Example. Integrating process with delay=1. G(s) = e -s /s. Model: k’=1,  =1,  1 =1 SIMC-tunings with  c with =  =1: IMC has  I =1 Ziegler-Nichols is usually a bit aggressive Setpoint change at t=0Input disturbance at t=20 TIGHT CONTROL

27 1.Approximate as first-order model with k=1,  1 = 1+0.1=1.1,  =0.1+0.04+0.008 = 0.148 Get SIMC PI-tunings (  c =  ): K c = 1 ¢ 1.1/(2¢ 0.148) = 3.71,  I =min(1.1,8¢ 0.148) = 1.1 2. Approximate as second-order model with k=1,  1 = 1,  2 =0.2+0.02=0.22,  =0.02+0.008 = 0.028 Get SIMC PID-tunings (  c =  ): K c = 1 ¢ 1/(2¢ 0.028) = 17.9,  I =min(1,8¢ 0.028) = 0.224,  D =0.22 TIGHT CONTROL

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29 Tuning for smooth control Will derive K c,min. From this we can get  c,max using SIMC tuning rule SMOOTH CONTROL Tuning parameter:  c = desired closed-loop response time Selecting  c =  (“tight control”) is reasonable for cases with a relatively large effective delay  Other cases: Select  c >  for  slower control  smoother input usage less disturbing effect on rest of the plant  less sensitivity to measurement noise  better robustness Question: Given that we require some disturbance rejection.  What is the largest possible value for  c ?  Or equivalently: The smallest possible value for K c ?

30 Closed-loop disturbance rejection d0d0 y max -d 0 -y max SMOOTH CONTROL =|u 0 |

31 KcKc u Minimum controller gain for PI-and PID-control: K c ¸ K c,min = |u 0 |/|y max | |u 0 |: Input magnitude required for disturbance rejection |y max |: Allowed output deviation SMOOTH CONTROL

32 Minimum controller gain: Industrial practice: Variables (instrument ranges) often scaled such that Minimum controller gain is then Minimum gain for smooth control ) Common default factory setting K c =1 is reasonable ! SMOOTH CONTROL (span)

33 Exampl e Does not quite reach 1 because d is step disturbance (not not sinusoid) Response to disturbance = 1 at input  c is much larger than  =0.25 SMOOTH CONTROL

34 Application of smooth control Averaging level control V q LC Reason for having tank is to smoothen disturbances in concentration and flow. Tight level control is not desired: gives no “smoothening” of flow disturbances. Let |u 0 | = |  q 0 | – expected flow change [m3/s] (input disturbance) |y max | = |  V max | - largest allowed variation in level [m3] Minimum controller gain for acceptable disturbance rejection: K c ¸ K c,min = |u 0 |/|y max | From the material balance (dV/dt = q – q out ), the model is g(s)=k’/s with k’=1. Select K c =K c,min. SIMC-Integral time for integrating process:  I = 4 / (k’ K c ) = 4 |  V max | / |  q 0 | = 4 ¢ residence time provided tank is nominally half full and  q 0 is equal to the nominal flow. SMOOTH CONTROLLEVEL CONTROL If you insist on integral action then this value avoids cycling

35 Rule: K c ¸ |u 0 |/|y max | =1 (in scaled variables) Exception to rule: Can have K c < 1 if disturbances are handled by the integral action.  Disturbances must occur at a frequency lower than 1/  I Applies to: Process with short time constant (  1 is small) and no delay (  ¼ 0).  Then  I =  1 is small so integral action is “large”  For example, flow control SMOOTH CONTROL K c : Assume variables are scaled with respect to their span

36 Summary: Tuning of easy loops Easy loops: Small effective delay (  ¼ 0), so closed- loop response time  c (>>  ) is selected for “smooth control” Flow control: K c =0.5,  I =  1 = time constant valve (typically, 10s to 30s) Level control: K c =2 (and no integral action) Other easy loops (e.g. pressure control): K c = 2,  I = min(4  c,  1 )  Note: Often want a tight pressure control loop (so may have K c =10 or larger) SMOOTH CONTROL K c : Assume variables are scaled with respect to their span

37 More on level control Level control often causes problems Typical story:  Level loop starts oscillating  Operator detunes by decreasing controller gain  Level loop oscillates even more ...... ??? Explanation: Level is by itself unstable and requires control. LEVEL CONTROL

38 Integrating process: Level control LEVEL CONTROL

39 How avoid oscillating levels? LEVEL CONTROL

40 Case study oscillating level We were called upon to solve a problem with oscillations in a distillation column Closer analysis: Problem was oscillating reboiler level in upstream column Use of Sigurd’s rule solved the problem LEVEL CONTROL

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42 Conclusion PID tuning

43 Cascade control

44 Tuning of cascade controllers

45 Cascade control serial process d=6 G1G1 u2u2 y1y1 K1K1 ysys G2G2 K2K2 y2y2 y 2s

46 Cascade control serial process d=6 Without cascade With cascade G1G1 u y1y1 K1K1 ysys G2G2 K2K2 y2y2 y 2s

47 Tuning cascade control: serial process Inner fast (secondary) loop:  P or PI-control  Local disturbance rejection  Much smaller effective delay (0.2 s) Outer slower primary loop:  Reduced effective delay (2 s instead of 6 s) Time scale separation  Inner loop can be modelled as gain=1 + 2*effective delay (0.4s) Very effective for control of large-scale systems

48 Controllability (Input-Output) “Controllability” is the ability to achieve acceptable control performance (with any controller) “Controllability” is a property of the process itself Analyze controllability by looking at model G(s) What limits controllability? CONTROLLABILITY

49 Recall SIMC tuning rules 1. Tight control: Select  c =  corresponding to 2. Smooth control. Select K c ¸ Must require K c,max > K c.min for controllability ) Controllability initial effect of “input” disturbance max. output deviation y reaches k’ ¢ |d 0 |¢ t after time t y reaches y max after t= |y max |/ k’ ¢ |d 0 | CONTROLLABILITY

50 Controllability CONTROLLABILITY

51 Example: Distillation column CONTROLLABILITY

52 Example: Distillation column CONTROLLABILITY

53 Distillation example: Frequency domain analysis CONTROLLABILITY

54 Other factors limiting controllability CONTROLLABILITY Input limitations (saturation or “slow valve”): Can sometimes limit achievable speed of response Unstable process (including levels): Need feedback control for stabilization ) Makes sure inputs do not saturate in stabilizing loops


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